FIRST DEGREE EQUATIONS

The mathematics lesson, equations subject. First degree equations in one variable. How to solve the first degree equations. The subject expression and example questions.

In the first degree equations that have a variable there is only one variable. The exponent of the variable is 1. Variables are called "unknown".

A first degree equation is shown below..

ax + b = 0

In the equation at above a and b are real number, x variable number.

A first degree equation is seen at below.

8x + 6 = 30

In the equation, x must atke a a value such that both sides of the equality are the same.

For x =1,

8.1 + 6 = 30

14 = 30→ it is wrong

For x = 2

8.2 + 6 = 30

22 = 30 → it is wrong

For x = 3,

8.3 +6 = 30

30 = 30 it is true

For x = 4

8.4 + 6 = 30

38 = 30 → it is wrong

Equation solution is operation that finding the unknown value.

We can apply at below steps to solve an equation.

1-The terms that contain the unknown value are collect at a side of the equality, other terms are collect at other side of the equality.

Example:

3x + 5 -7 -4x = 8 – 6x + 14 + x

Find the value of x in the equation above.

Solution:

Let we carry the x contain terms to the left side of the equation, constant values to the right side.

When a term is carried from a side of the equality to the other side of the equality, the sign of the term changes.

3x – 4x + 6x – x = 8 + 14 – 5 + 7

4x = 24

x = 24/4

x = 6

2- Rational terms are collected among themselves.

If there is rational terms at both sides of the equation, the denominator of the left side of the equation multiplied by the numerator of the right side of the equation. The numerator of the left side of the equation is multiplied by the denominator of the right side of the equation.

Example:

Find value of x in the equation below.

3x – 4

+ 2x + 3 = 5 –

3-x

+ 9

2

3

Solution:

Step -1

Let we take the terms that contain x to the left side of equation.

3x – 4

+ 2x +

3 – x

= 5 + 9 - 3

2

3

Step -2

Addition the terms at the both sides.

6x – 8 + 9 – 3x

+ 2x = 11

6

3x + 1 + 12x

= 11

6

Step -3

Now, let we multiply the right side by the denominator of the left side.

15x + 1 =6.11

This operation is equal to operation below. So, we are multiplying the both sides of the equality by 6.

6.

15x + 1

= 6.11

6

15x + 1 = 66

15x = 65

3x = 13

x = 13/3

Example:

Find value of x in the equation below.

3 – 4x

+

5

+ 2x =

7

+ 3 –

x

2

3

3

3

Solution:

Let we carry the x contain terms to the left side of the equation, constant values to the right side.

3 – 4x

+ 2x +

x

=

7

+ 3 -

5

3

3

2

3

6 – 8x

+

3x

+ 2x =

7 - 5

+ 3

3

6

6

6 – 8x + 3x

+ 2x =

2

+ 3

3

6

6 – 5x + 12x

=

11

3

6

7x + 6

=

11

3

6

Reciprocal, the denominators is multiplied by the shares, and the shares is multiplied by denominators.

21x + 18 = 66

21x = 48

x =

48

21

x =

16

7

3- If the unknown x variable is in the root, is taken square of the both sides.

Example:

Find the x value in the equation below.

√5x + 1 – 6 = 0

Solution:

let we carry the 6 to the right side of the equality.

√5x + 1 = 6

Let we take the square of the both sides of the equality.

(√5x + 1)2 = 62

5x + 1 = 36

5x = 35

x = 7

4- If there is exponent at both sides of the equality, the exponents are simplification.

Example:

Find unknown value in the equation below.

(3 – 2x)3 = (x + 1)3

Solution:

Let we simplify the mutually exponents.

3 – 2x = x + 1

2 = 3x

3x = 2

x = 2/3

The first degree equations related questions and the solutions of these questions are take place at the link below.

The absolute value equations and the exponential quations will be examined under a seperate heading.